How do you calculate $arcsin (- \frac{1}{\sqrt{2}})$ $arcsin (-1/ sqrt 2)$ ?

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How do you calculate $arcsin (- \frac{1}{\sqrt{2}})$ $arcsin (-1/ sqrt 2)$ ?

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Answer 1 · 2015-07-20T21:14:43

Use properties of right angled triangle with sides $\frac{1}{\sqrt{2}}$ $1/sqrt(2)$ , $\frac{1}{\sqrt{2}}$ $1/sqrt(2)$ and $1$ $1$ to find $arcsin (- \frac{1}{\sqrt{2}}) = - \frac{π}{4}$ $arcsin(-1/sqrt(2)) = -pi/4$

Explanation:

$arcsin (- \frac{1}{\sqrt{2}}) = α$ $arcsin(-1/sqrt(2)) = alpha$ where $α \in [- \frac{π}{2}, \frac{π}{2}]$ $alpha in [-pi/2, pi/2]$ such that $sin (α) = - \frac{1}{\sqrt{2}}$ $sin(alpha) = -1/sqrt(2)$

$\frac{1}{\sqrt{2}}$ $1/sqrt(2)$ is one side of the right angled triangle with sides:

$\frac{1}{\sqrt{2}}$ $1/sqrt(2)$ , $\frac{1}{\sqrt{2}}$ $1/sqrt(2)$ and $1$ $1$

(Note ${(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = \frac{1}{2} + \frac{1}{2} = 1 = 1^{2}$ $(1/sqrt(2))^2 + (1/sqrt(2))^2 = 1/2 + 1/2 = 1 = 1^2$ )

This triangle has angles $\frac{π}{4}$ $pi/4$ , $\frac{π}{4}$ $pi/4$ and $\frac{π}{2}$ $pi/2$ .

So $sin (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $sin(pi/4) = 1/sqrt(2)$

Now $sin (- θ) = - sin (θ)$ $sin(-theta) = -sin(theta)$

So $sin (- \frac{π}{4}) = - \frac{1}{\sqrt{2}}$ $sin(-pi/4) = -1/sqrt(2)$

So $arcsin (- \frac{1}{\sqrt{2}}) = - \frac{π}{4}$ $arcsin(-1/sqrt(2)) = -pi/4$

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How do you calculate $arcsin (- \frac{1}{\sqrt{2}})$ $arcsin (-1/ sqrt 2)$ ?

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